Upload file using python requests

问题

I've been trying to upload a file using the box v2 api with requests.

So far I had little luck though. Maybe someone here can help me see what I'm actually doing wrong.

file_name = "%s%s" % (slugify(sync_file.description), file_suffix)
file_handle = open(settings.MEDIA_ROOT + str(sync_file.document), 'rb')
folder_id = str(sync_file.patient.box_patient_folder_id)

r = requests.post(
    files_url,
    headers=headers,
    files={
        file_name: file_handle,
        "folder_id": folder_id,
    },
)

My authentication works, because I'm creating a folder just before that, using the same data.

A response looks something like this:

{
    u'status': 404, 
    u'code': u'not_found', 
    u'help_url': u'http://developers.box.com/docs/#errors', 
    u'request_id': u'77019510950608f791a0c1', 
    u'message': u'Not Found', 
    u'type': u'error'
}

Maybe someone on here ran into a similar issue.

回答1:

You need to pass 2 Python dictionaries, files and data. files are {uniqFileName:openFileObj}, and data are {uniqFileName:filename}. Below is the upload method from my box class. And remember to add a final entry in data, 'folder_id': destination_id.

def uploadFiles(self, ufiles, folid):
    '''uploads 1 or more files in the ufiles list of tuples containing
    (src fullpath, dest name). folid is the id of the folder to
    upload to.'''

    furl = URL2 + 'files/data'
    data, files = {}, {}
    for i, v in enumerate(ufiles):
        ff = v[0]
        fn = v[1]
        #copy to new, renamed file in tmp folder if necessary
        #can't find a way to do this with the api
        if os.path.basename(ff) != fn:
            dest = os.path.join(TMP, fn)
            shutil.copy2(ff, dest)
            ff = dest

        f = open(ff, 'rb')
        k = 'filename' + str(i)
        data[k] = fn
        files[k] = f

    data['folder_id'] = folid

    res = self.session.post(furl, files=files, data=data)

    for k in files:
        files[k].close()


    return res.status_code

Here is a sample call:

destFol = '406600304'

ret = box.uploadFile((('c:/1temp/hc.zip', 'hz.zip'),), destFol)

Like I said, the above function is a method of a class, with an instance attr that holds a requests session. But you can use requests.post instead of self.session.post, and it will work just the same. Just remember to add the headers with your apikey and token if you do it outside a session.

According to the documentation, you are supposed to be able to rename the file by giving it a new name in the data dict. But I can't make this work except by copying the src file to a temp dir with the desired name and uploading that. It's a bit of a hack, but it works.

good luck, Mike

回答2:

As someone requested my implementation, I figured I would put it out here for anyone trying to achieve something similar.

files_url = "%s/files/content" % (settings.BOX_API_HOST)
headers = {"Authorization": "BoxAuth api_key=%s&auth_token=%s" % 
              (settings.BOX_API_KEY, self.doctor.box_auth_token)
          }

file_root, file_suffix = os.path.splitext(str(self.document))
filename = "%s%s" % (slugify(self.description), file_suffix)
files = {
        'filename1': open(settings.MEDIA_ROOT + str(self.document), 'rb'),
        }
data = {
        'filename1': filename,
        'folder_id': str(self.patient.get_box_folder()),
       }

r = requests.post(files_url,
                  headers=headers,
                  files=files,
                  data=data)

file_response = simplejson.loads(r.text)

try:
    if int(file_response['entries'][0]['id']) > 0:
        box_file_id = int(file_response['entries'][0]['id'])

        #Update the name of file
        file_update_url = "%s/files/%s" % (settings.BOX_API_HOST, box_file_id) 
        data_update = {"name":  filename}
        file_update = requests.put(file_update_url,
                                   data=simplejson.dumps(data_update),
                                   headers=headers)

        LocalDocument.objects.filter(id=self.id).update(box_file_id=box_file_id)
except:
    pass

So in essence, I needed to send the file and retrieve the ID of the newly updated file and send another request to box. Personally, I don't like it either, but it works for me and haven't been able to find any other implementations that do the correct naming from the get-go.

Hope someone can benefit from this snippet.

回答3:

My solution, using requests:

def upload_to_box(folder_id, auth_token, file_out):
    headers = { 'Authorization' : BOX_AUTH.format(auth_token) }
    url = 'https://api.box.com/2.0/files/content'
    files = { 'filename': (new_file_name, open(file_out,'rb')) }
    data = { 'folder_id': folder_id }
    response = requests.post(url, params=data, files=files, headers=headers)

It would be nice if you could specify the new_copy parameter but there's nothing documented for it and it doesn't seem to work.

来源：https://stackoverflow.com/questions/12569545/upload-file-using-python-requests

标签

box-api

python-requests